If `phi_(1) and phi_(2)` are normal of flux entering and leaving an enclosed surface. Electric flux inside the surface will be

Options

A. `(phi_(2)- phi_(1))epsi_(0)`
B. `(phi_(1) + phi_(2)) epsi_(0)`
C. `((phi_(2)-phi_(1)))/(epsi_(0))`
D. `((phi_(2) + phi_(1)))/(epsi_(0))`

Explanation

To solve the problem regarding the electric flux inside an enclosed surface, we can use Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space.

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
Electric flux (Φ) through a surface is defined as the product of the electric field (E) and the area (A) through which it passes, taking into account the angle (θ) between the field lines and the normal to the surface. Mathematically, it is given by:
\[ \Phi = E \cdot A \cdot \cos(\theta) \]

2. **Applying Gauss's Law**:
According to Gauss's Law, the total electric flux (Φ_total) through a closed surface is related to the charge (Q) enclosed by that surface:
\[ \Phi_{\text{total}} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]
where \( \varepsilon_0 \) is the permittivity of free space.

3. **Flux Entering and Leaving the Surface**:
If \( \Phi_1 \) is the flux entering the surface and \( \Phi_2 \) is the flux leaving the surface, the net flux (Φ_net) through the surface can be expressed as:
\[ \Phi_{\text{net}} = \Phi_1 - \Phi_2 \]

4. **Electric Flux Inside the Surface**:
The electric flux inside the surface is determined by the net flux. If there is no charge inside the surface, then \( Q_{\text{enc}} = 0 \), leading to:
\[ \Phi_{\text{net}} = 0 \implies \Phi_1 = \Phi_2 \]
Hence, the electric flux inside the surface will be zero if the entering and leaving fluxes are equal.

5. **Conclusion**:
Therefore, the electric flux inside the enclosed surface can be concluded as:
\[ \Phi_{\text{inside}} = \Phi_1 - \Phi_2 \]

### Final Answer:
The electric flux inside the surface will be \( \Phi_{\text{inside}} = \Phi_1 - \Phi_2 \).

---